Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
int t;
int main() {
  cout.sync_with_stdio(false);
  cin.tie(NULL);
  cout.setf(ios_base::fixed);
  cout.precision(2);
  int a, b;
  cin >> a >> b;
  vector<vector<int> > v(a, vector<int>(b)), x;
  for (int i = 0; i < a; ++i)
    for (int j = 0; j < b; ++j) cin >> v[i][j];
  vector<pair<int, int> > r1, r2;
  bool flag = true, flag2 = true;
  x = v;
  for (int i = 0; i < a; ++i) {
    int mx = 1e9;
    for (int j = 0; j < b; ++j) mx = max(min(mx, v[i][j]), 0);
    int k = mx;
    while (k-- > 0) r1.push_back(pair<int, int>(i, 0));
    for (int j = 0; j < b; ++j) {
      int h = max(x[i][j] - mx, 0);
      for (int k = 0; k < a; ++k) x[k][j] -= h;
      while (h-- > 0) r1.push_back(pair<int, int>(j, 1));
      x[i][j] -= mx;
    }
  }
  for (auto& i : x)
    for (auto& j : i)
      if (j < 0) flag = false;
  x = v;
  for (int i = 0; i < b; ++i) {
    int mx = 1e9;
    for (int j = 0; j < a; ++j) mx = max(0, min(mx, v[j][i]));
    int k = mx;
    while (k-- > 0) r2.push_back(pair<int, int>(i, 1));
    for (int j = 0; j < a; ++j) {
      int h = max(x[j][i] - mx, 0);
      for (int k = 0; k < b; ++k) x[j][k] -= h;
      while (h-- > 0) r2.push_back(pair<int, int>(j, 0));
      x[j][i] -= mx;
    }
  }
  for (auto& i : x)
    for (auto& j : i)
      if (j < 0) flag2 = false;
  if (!flag && !flag2)
    cout << "-1" << '\n';
  else {
    vector<pair<int, int> > res;
    if (flag && flag2) {
      if (r1.size() < r2.size())
        res = r1;
      else
        res = r2;
    } else if (flag)
      res = r1;
    else
      res = r2;
    cout << res.size() << '\n';
    for (pair<int, int> p : res)
      cout << (p.second ? "col " : "row ") << p.first + 1 << '\n';
  }
  return 0;
}